Answer
-1.79 V
Work Step by Step
The reaction at the cathode is the reduction of $Ni^{2+}$ to $Ni$ and the reaction at the anode is the oxidation of $U$ to $U^{3+}$.
$E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}$
$\implies E^{\circ}_{anode}=E^{\circ}_{cathode}-E^{\circ}_{cell}$
Or $E^{\circ}_{U^{3+}/U}=E^{\circ}_{Ni^{2+}/Ni}-E^{\circ}_{cell}=-0.25\,V-1.54\,V=-1.79\,V$
